The minimum axial compressive load, P, required to initiate buckling for a pinned-pinned slender column with bending stiffness EI and length L is

The correct option is the second option, which is represented by the formula shown in the image image_1.png:
$$P=\frac{{\pi }^{2}EI}{L^2}$$

Step-by-Step Derivation and Explanation:

1. Understanding the Setup:
Consider a slender column of length L and bending stiffness EI, under an axial compressive load P.
The column is "pinned-pinned" (or simply supported) at both ends. This means that both ends are constrained against lateral displacement but are free to rotate.

2. Governing Differential Equation:
According to Euler's buckling theory, if the column deflects laterally by a distance y(x) at any point x along its length, the internal bending moment at that section is:
$$M\left(x\right)=-P·y$$
Using the Euler-Bernoulli beam theory, the relationship between the bending moment and deflection is:
$$EI\frac{d^{2}y}{d{x}^2}=M\left(x\right)$$
Substituting the bending moment equation into the beam equation gives the governing second-order differential equation for buckling:
$$EI\frac{d^{2}y}{d{x}^2}+Py=0$$

3. General Solution:
Let us define a constant parameter:
$$k^2=\frac{P}{EI}$$
Using this definition, the differential equation simplifies to:
$$\frac{d^{2}y}{d{x}^2}+k^{2}y=0$$
The general solution to this ordinary differential equation is:
$$y\left(x\right)=Acos\left(kx\right)+Bsin\left(kx\right)$$
where A and B are constants determined by the boundary conditions of the column.

4. Applying Boundary Conditions:
For a pinned-pinned column, the lateral displacement y must be zero at both supported ends:
- At the bottom support, where:
$$x=0$$
We apply the boundary condition:
$$y\left(0\right)=Acos\left(0\right)+Bsin\left(0\right)=0$$
Since:
$$cos\left(0\right)=1\text{and}sin\left(0\right)=0$$
We get:
$$A=0$$
This simplifies the deflection equation to:
$$y\left(x\right)=Bsin\left(kx\right)$$
- At the top support, where:
$$x=L$$
We apply the boundary condition:
$$y\left(L\right)=Bsin\left(kL\right)=0$$

5. Finding the Critical Buckling Load:
For buckling to occur, the column must experience a lateral deflection, which means we must find a non-trivial solution where:
$$B\ne 0$$
Therefore, we must satisfy the condition:
$$sin\left(kL\right)=0$$
This sinusoidal equation is satisfied when:
$$kL=n\pi$$
where n is a positive integer:
$$n=1,2,3,...$$
Each value of n corresponds to a specific buckling mode shape.

6. Minimum Load to Initiate Buckling:
The minimum compressive load required to initiate buckling occurs at the lowest energy state, which corresponds to the first buckling mode:
$$n=1$$
This gives:
$$kL=\pi \Rightarrow k=\frac{\pi }{L}$$
Squaring both sides:
$$k^2=\frac{{\pi }^2}{L^2}$$
Substituting our definition of $k^2$ back into the equation:
$$\frac{P}{EI}=\frac{{\pi }^2}{L^2}$$
Solving for the minimum axial buckling load P:
$$P=\frac{{\pi }^{2}EI}{L^2}$$
This matches the formula presented in image_1.png.